The identity below is significant because it relates 3 different kinds of products: a cross product and a dot product of 2 vectors on the left side, and the product of 2 real numbers on the right side. Prove the identity below.
| a × b |² + (a • b)² = |a|²|b|²
My work, LSH:
= | a × b |² + (a • b)²
= (|a||b|sinθ)(|a||b|sinθ) + (|a||b|cosθ)(|a||b|cosθ)
= (|a|²)(|a||b|)(|a|sinθ)(|a||b|)(|b|²)(|b|sinθ)(|a| sinθ)(|b|sinθ)(sin²θ) + (|a|²)(|a||b|)(|a|cosθ)(|a||b|)(|b|²)(|b|cosθ)(|a| cosθ)(|b|cosθ)(cos²θ)
= (|a|²)(|a||b|)²(|a|sinθ)²(|b|²)(|b|sinθ)²(sin²θ) + (|a|²)(|a||b|)²(|a|cosθ)²(|b|²)(|b|cosθ)²(cos²θ)
= (|a|²|b|²(|a||b|)²) [(|a|sinθ)²(|b|sinθ)²(sin²θ) + (|a|cosθ)²(|b|²)(|b|cosθ)²(cos²θ)]
= (|a|²|b|²(|a||b|)²) [(|a|²)(sin²θ)(|b|²)(sin²θ)(sin²θ) + (|a|²)(cos²θ)(|b|²)(|b|²)(cos²θ)(cos²θ)]
= (|a|²|b|²(|a||b|)²) [(sin²θ)(sin²θ)(sin²θ) + (cos²θ)(cos²θ)(cos²θ)]
And now I don't know what else to do! Please help. Did I mess up somewhere in my steps? Or is it possible to common factor still?
Hi there,
= | a × b |² + (a • b)²
= (|a||b|sinθ)(|a||b|sinθ) + (|a||b|cosθ)(|a||b|cosθ)
it's good up to here but then you're problem was how you multiplied out the brackets, remember multiplication distributes over addition, so if you had
(|a|+|b|+sinθ)(|a|+|b|+sinθ) you would have been multiplying them out the correct way (ish)... however you just have a product, so just stick them together and carry on, the last step given by topsquark follows from the trigonometric identity cos^2 +sin^2=1 which can be proved using pythagoras theorem and the unit circle (ask me or search the web if you are interested..)